I have a quadratic bezier curve and I want to calculate the slope of the tangent in a given point. For example, let it be the middlepoint of the quadratic bezier curve, therefore t=0.5 (please see the link below for a picture of this). I've calculated the first derivative of the formula for the quadratic bezier curve; however I get 400 as value for the slope, though it should be 0. Maybe I'm using the first derivative in a wrong way? I know I could also calculate the tangents using trigonometric functions; however I'd like to do it using the first derivative, shouldn't this be possible? Thanks for any hint!
For clarification / please note: I'm interested in a general way to get the slope in a arbitrary given point on a quadratic bezier curve, not only to get the tangent in the start- and end point.
A picture of my problem including the text above:
http://cid-0432ee4cfe9c26a0.skydrive.live.com/self.aspx/%c3%96ffentlich/Quadratic%20Bezier%20Curve.pdf
Thank you very much for any hint!
Using your formula for B'(t), evaluated at t=1/2, we get
B'(1/2) = -P0 + P2
From the look of your graph, P0 = (0,0) and P2 = (400,0). So
B'(1/2) = (400,0).
This is the "velocity" of a point traveling along B(t) at t=1/2.
(400,0) is a horizontal vector, with magnitude 400.
So all is as it should be. Since B'(t) is horizontal, it does have "slope" 0.
Derivatives of a Bézier Curve
Related
For a cubic bezier curve, if its parameter t is 0.5 then does this split the curve equal pieces lengthwise? Reasoning is appreciated.
In general, no, unless in some special cases, for example all control points of the cubic Bezier curve are collinear and evenly-spaced. The reason is simple: a Bezier curve is in general not arc-length parameterized.
Simple example: take three 2D control points
(1,0), (2,0) and (4,0)
The point t=0.5 is (2,0). But this point does not have equal distances to the first and last point.
Well, and this should not depend on the cubic or linear case.
So I'm trying to animate a sprite, and I just found the CCBezierBy/CCBezierTo methods, which are saving me a lot of trouble, but they only work as cubic functions.
HOw can I make them quadratics? (I need s-curves)
Having only 3 controlpoints limits severely what you can do.
Thank you.
I don't think you understand Bezier curves here: S curves cannot be modeled by quadratic Beziers, they can only do curves with a single inflection point. Cubic curves can have two inflection points, so quadratic curves can only model curves that look like | or C, and cubic curves can model anything that looks like |, C or S
Cubic functions have a starting on-curve point, two control points, and an ending on-curve point, for a total of four control points; Quadratic functions have a starting on-curve point, one control point, and an ending on-curve point, for a total of three control points.
So what you want is a cubic curve, and that's exactly what the CCBezierBy/CCBezierTo methods offer. They let you specify a bezier curve as the two control points and end point, because you'll already have the starting point.
This is my first time posting here, so I hope this is ok. I'm working on a java project but my question is really about the math I'll be using for it...
I have three (different) points at (x1, y1), (x2, y2), and (x3, y3). All I need is a formula for the length of the cubic spline formed between them. For someone good at calculus, this should be pretty easy to derive. I've looked all around online but can't seem to find the solution. Again, I don't even need the equation of the spline - just its length, given the three points. Thanks in advance! If someone can figure this out and share, you'll makey day :)
I have some bad news.
The first is that a cubic b-spline generally takes 4 points to define. It is possible to define one from 3 points, but it usually involves making up another point somehow (for example, using degree elevation). So we'd need to have information about how exactly you're defining the spline - if it's some other kind of spline (catmull-rohm?), or the details of how you're constructing it.
The second is that there's no closed-form equation for the length of a b-spline, or even a Bezier curve. What I typically do is sample the curve at a lot of points, and then compute the length of the polyline.
There are formulas that can tell you what your error bound will be, based on the derivatives of the curve, and there are methods that approximate using arcs rather than line segments, but those are probably more complicated than they're worth.
See the primer on bezier curves for a more info. However, sadly, tfinniga is correct for cubic splines you need to use an approximation.
Considering the following nice solution for finding cubic Bézier control points for a curve passing through 4 points:
How to find control points for a BezierSegment given Start, End, and 2 Intersection Pts in C# - AKA Cubic Bezier 4-point Interpolation
I wonder, if there is a straightforward extension to this for making the Bézier curve pass through N points, for N > 2 and maybe N ≤ 20?
This is a really old question, but I'm leaving this here for people who have the same question in the future.
#divanov has mentioned that there's no Bezier curve passing through N arbitrary points for N >4.
I think the OP was asking how to compute the control points to join multiple bezier curves to produce a single curve that looks smooth.
This pdf will show you how to compute the control points: http://www.math.ucla.edu/~baker/149.1.02w/handouts/dd_splines.pdf
which I found on this writeup https://developer.squareup.com/blog/smoother-signatures/ from Square about how they render a smooth curve that passes through all the sampled points of a mouse drawn signature.
In general, there is no Bezier curve passing through N arbitrary points, where N > 4. One should consider curve fitting to minimize least square error between computed Bezier curve and given N data points. Which is discussed, for example, here.
I want to smooth some hand draw lines in iphone.
I have used the following code in
http://webdocs.cs.ualberta.ca/~graphics/books/GraphicsGems/gems/FitCurves.c
However, I found that some bezier curved was wrong, the second control point and end point is invalid.
Did anybody have the same problem before?
Thanks.
Bezier Curves are not designed to go through the provided vertices!
They are designed to shape a smooth curve influenced by the control points.
First you must decide if you want to interpolate between missing points,
or if you want to filter non smooth data:
Filtering
You should look at "sliding average" with a small averaging window. (try 5 - 10 pixel).
This works as follows: (look for wiki for a detailed description)
I use here an average window of 10 points:
start by calculation of the average of points 0 - 9, and output the result as result point 0
then calculate the average of point 1 - 10 and output, result 1
And so on.
Interpolation
If you want to interpolate between (missing) points using a smooth curve, you could use piece - wise cubic splines:
You calculate the coefficients of a cubic polygon through 3 vertices.
You start with calculating the cubic polygon through:
Point[0] - Point[2], but you draw your output only from Point[0] to Point[1] .
Then you move on one step: and calculate through
Point[1] - Point[3], but you draw only from p1 to p2.
And so on.
You need to search on wiki for cubic interpolation for a detailed explanation how to caluclate a cubic polygon (sometimes called cubic spline).